A stone thrown vertically upwards with initial velocity $u$ reaches a height $^{'} h^{'}$ before coming down. Show that the time taken to go up is same as the time taken to come down.

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Solution

(i) When stone throw upward with initial velocity, $u$

Final velocity at maximum height, $v = 0$

Apply kinematic equation of motion.

$v = u + a t$

$t = \frac{v - u}{a} = \frac{0 - u}{- g}$

Apply kinematic equation of motion.

$s = u t + \frac{1}{2} a t^{2}$

$h = u (\frac{u}{g}) - \frac{1}{2} g {(\frac{u}{g})}^{2}$

$u = \sqrt{2 g h}$

Time,

$t = \frac{u}{g} = \frac{\sqrt{2 g h}}{g}$

$t = \sqrt{\frac{2 h}{g}} . . . . . . . . (1)$

(ii) When stone drop from height h$$

Apply kinematic equation of motion.

$v^{2} - u^{2} = 2 g h$

$v = \sqrt{2 g h}$

Apply kinematic equation of motion.

$v = u + a t$

$t = \frac{v - u}{a} = \frac{\sqrt{2 g h} - 0}{g} = \sqrt{\frac{2 h}{g}} . . . . . . . (2)$

From (1 ) and (2) it is proved that the time taken to go up is same as the time taken to come down.

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A stone thrown vertically upwards with initial velocity u reaches a height ′h′ before coming down. Show that the time taken to go up is same as the time taken to come down.

A stone thrown vertically upwards with initial velocity $u$ reaches a height $^{'} h^{'}$ before coming down. Show that the time taken to go up is same as the time taken to come down.